State and prove Newton’s law of cooling. Explain how it could be verified experimentally
Answers
Answer:
It is relatively easy to experimentally verify Newton's Law of Cooling. As mentioned above, the Law of Cooling states that ¨the rate at which a warm body cools is proportional to the difference between the temperature of the warm body and the temperature of its environment.
Explanation:
Answer:
Statement:
According to Newton’s law of cooling, the rate of loss of heat from a body is directly proportional to the difference in the temperature of the body and its surroundings.
Derivation:
For small temperature difference between a body and its surrounding, the rate of cooling of the body is directly proportional to the temperature difference and the surface area exposed.
dQ/dt ∝ (q – qs)], where q and qs are temperature corresponding to object and surroundings.
From above expression , dQ/dt = -k[q – qs)] . . . . . . . . (1)
This expression represents Newton’s law of cooling. It can be derived directly from stefan’s law, which gives,
k = [4eσ×θ3o/mc] A . . . . . (2)
Now, dθ/dt = -k[θ – θo]
⇒ \int_{\theta_1}^{\theta_2}\frac{d\theta}{(\theta-\theta_o)} = \int_{0}^{1}-k dt∫
θ
1
θ
2
(θ−θ
o
)
dθ
=∫
0
1
−kdt
Newton's Law of Cooling Derivation
where,
qi = initial temperature of object,
qf = final temperature of object.
ln (qf – q0)/(qi – q0) = kt
(qf – q0) = (qi – q0) e-kt
qf = q0 + (qi – q0) e -kt . . . . . . (3).
⇒ Check: Heat transfer by conduction
Experiment:
Water is heated to 80oC for 10 min. How much would be the temperature if k = 0.056 per min and the surrounding temperature is 25oC?
Solution:
Given:
Ts = 25oC,
To = 80oC,
t = 10 min,
k = 0.056
Now, substituting the above data in Newton’s law of cooling formula,
T(t) = Ts + (To – Ts) × e-kt
= 25 + (80 – 25) × e-0.56 = 25 + [55 × 0.57] = 45.6 oC
Temperature cools down from 80oC to 45.6oC after 10 min.